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I was given an excersice to solve wich asks you to prove:

$$\int_0^1 f(r) r dr = 0 $$

knowing that:

$$\int_0^1 f(t) dt = 0 $$

After doing integration by parts I ended up with:

$$\int f(r) r dr = (\int f(r)dr)r- \int\int f(r)drdr $$

My question is, how should I proceed in order to evaluate the integral with my integration limits between 0 and 1. I tried with Barrow, but I don't know what to do with the double integral and its limis

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  • $\begingroup$ "I was given an exercise to solve... " Where is this exercise from? $\endgroup$
    – user587192
    Dec 11, 2018 at 18:57
  • $\begingroup$ "I don't know what to do with the double integral and its limis" No, you don't have a double integral. $\endgroup$
    – user587192
    Dec 11, 2018 at 18:58
  • $\begingroup$ Suppose $F'(x)=f(x)$. Integration by parts says that $$ \int_0^1f(x)x\ dx = F(x)x\mid_{0}^1-\int_0^1F(x)dx. $$ $\endgroup$
    – user587192
    Dec 11, 2018 at 19:02

2 Answers 2

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You can't prove it, since it is false. If $f(x)=\sin(2\pi x)$, then $\displaystyle\int_0^1f(x)\,\mathrm dx=0$, but $\displaystyle\int_0^1x f(x)\,\mathrm dx=-\frac1{2\pi}\neq0$

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  • $\begingroup$ I guess the one who wrote the exercise forgot to add "prove, if possible". Thank you a lot! $\endgroup$
    – Alexander
    Dec 11, 2018 at 19:14
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It does not work for a linear function: $$\int_0^1 \left(x-\frac12\right)dx=\left(\frac{x^2}{2}-\frac x2\right)|_0^1=0, \ \ \text{but}\\ \int_0^1 \left(x-\frac12\right)xdx=\left(\frac {x^3}{3}-\frac{x^2}{4}\right)|_0^1=\frac1{12}\ne 0.$$

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